762 CHAPTER 11 Systems of Equations and Inequalities Recall that a solution to a system of equations consists of values for the variables that are solutions of each equation of the system. For example, = x 3, = − = − y z 1, 5 is a solution to the system of equations ( ) ( ) ( ) ( ) ( ) + + = − − + =− − + =− ⎧ ⎨ ⎪⎪ ⎪⎪ ⎩ ⎪⎪ ⎪⎪ + − + − = − ⋅−−+−=+− =− − ⋅ + − = − − = − ⎧ ⎨ ⎪⎪ ⎪ ⎩ ⎪⎪ ⎪⎪ x y z x y z x y 3 2 3 6 21 3 5 14 3 1 5 3 2331 65 6330 21 3 3 5 1 9 5 14 (1) (2) (3) because these values of the variables are solutions of each equation. Typically, when solving a system of three linear equations containing three variables, we use the method of elimination. Recall that the idea behind the method of elimination is to form equivalent equations so that adding two of the equations eliminates a variable. Solving a System of Three Linear Equations with Three Variables Use the method of elimination to solve the system of equations. + − =− − + = − − = ⎧ ⎨ ⎪⎪ ⎪⎪ ⎩ ⎪⎪ ⎪⎪ x y z x y z x y z 1 4 3 2 16 2 2 3 5 (1) (2) (3) Solution EXAMPLE 9 For a system of three equations, attempt to eliminate one variable at a time, using pairs of equations, until an equation with a single variable remains. Our strategy for solving this system will be to use equation (1) to eliminate the variable x from equations (2) and (3). We can then treat the new equations (2) and (3) as a system with two variables.Alternatively, we could use equation (1) to eliminate either y or z from equations (2) and (3). Try one of these approaches for yourself. We begin by multiplying both sides of equation (1) by −4 and adding the result to equation (2). (Do you see why? The coefficients of x are now opposites of one another.) We also multiply equation (1) by −2 and add the result to equation (3). Notice that these two procedures result in the elimination of the variable x from equations (2) and (3). + − =− − + = ⎧ ⎨ ⎪⎪ ⎩ ⎪⎪ x y z x y z 1 4 3 2 16 (1) Multiply by 4. − (2) x y z x y z y z 4 4 4 4 4 3 2 16 7 6 20 − − + = − + = ⎧ ⎨ ⎪⎪ ⎩ ⎪⎪ − + = (1) (2) Add. + − =− − − = ⎧ ⎨ ⎪⎪ ⎩ ⎪⎪ x y z x y z 1 2 2 3 5 (1) Multiply by 2. − (3) x y z x y z y z 2 2 2 2 2 2 3 5 4 7 − − + = − − = ⎧ ⎨ ⎪⎪ ⎩ ⎪⎪ − − = (1) (3) Add. x y z y z y z 1 7 6 20 4 7 + − =− − + = − − = ⎧ ⎨ ⎪⎪ ⎪⎪ ⎩ ⎪⎪ ⎪⎪ (1) (2) (3) Now concentrate on the new equations (2) and (3), treating them as a system of two equations containing two variables. It is easier to eliminate z. Multiply equation (3) by 6, and add equations (2) and (3). − + = − − = y z y z 7 6 20 4 7 (2) (3) Multiply by 6. y z y z y 7 6 20 24 6 42 31 62 − + = − − = − = (2) (3) Add. x y z y z y 1 7 6 20 31 62 + − =− − + = − = ⎧ ⎨ ⎪⎪ ⎪⎪ ⎩ ⎪⎪ ⎪⎪ (1) (2) (3) Now solve the new equation (3) for y by dividing both sides of the equation by −31. x y z y z y 1 7 6 20 2 + − =− − + = =− ⎧ ⎨ ⎪⎪ ⎪⎪ ⎩ ⎪⎪ ⎪⎪ (1) (2) (3) Back-substitute = − y 2 in equation (2) and solve for z. y z z z z 7 6 20 7 2 6 20 6 6 1 ( ) − + = − − + = = = (2) Substitute y 2. =− Subtract 14 from both sides of the equation. Solve for z. COMMENT Most handheld graphing calculators cannot graph in three dimensions. Web-based tools are available that can graph in space. ■
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